GATE CSE 2006 | Question: 09, ISRO2009-35

Question

A CPU has $24$-$bit$ instructions. A program starts at address $300$ (in decimal). Which one of the following is a legal program counter (all values in decimal)?

• A. $400$
• B. $500$
• C. $600$
• D. $700$

Answer 1

Option (C) will be the answer because here because here 24 bits = 3 bytes. Assuming system to be byte addressable the address of the instructions will increase by 3 bytes or 6 bytes or 9 bytes or.... etc for continuous instructions. so as 300 is multiple of 3 and as subsequent addresses are also multiple of 3 the answer will be 600 because it is the only multiple of 3 present in the options.

Answer 2

C.P.U will use 24- bit instructions and instructions size(bytes) is 24/8 = 3 bytes

Program execution will start from address at 300 and from there on words every operation will takes 3- bytes and so on , So the address should be multiple of “3”. From the given options only 600 address value is valid.

Answer 3

Yes, PC will have multiples of 3.

*as we considered the system as byte addressable, each instruction contains 3 bytes, so PC will increment by 3.

But, for this reason we can't just say that the answer would be multiple of 3. This might not be true all the time.

Here it worked because the starting address is multiple of 3. (300+answer) should be multiple of 3.

What if the starting address is not a multiple of 3.

In that case, we must subtract the starting address from all the options, the result which is multiple of 3 would be the answer.

So, (Option - starting_address) must be multiple of 3.

(600 - 300) = 300, is a multiple of 3.

# almost all the given answers are correct, but mine is an extension, where the starting address is not a multiple of 3.